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Proof step 1 from proof i2b-equal (conclusion)

⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: byte-equal: ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;